PROJECTILE MOTION Chapter 3, sections 3.1-3.3, 3.5 EXAMPLES OF PROJECTILE MOTION PROJECTILE MOTION vx v vx: Horizontal Velocity. Once in air, vx does not change since there is no force pushing or pulling horizontally. vy: Vertical Velocity. Once in air, vy changes because of the force of gravity. The projectile accelerates at the rate of 9.8 m/s2 downward. v PM simulation PROJECTILE MOTION • A projectile is an object shot through the air upon which the only force acting is gravity (neglecting air resistance). • The trajectory (path through space) or a projectile is parabolic. Time of flight (for horizontally launched projectile): y v0 yt 1 / 2 a yt 1 / 2a yt t 2 2 2y ay Time it takes to hit the ground is completely unaffected by the horizontal velocity The horizontal part of a projectile motion problem is a constant velocity problem. • • The vertical part of a projectile motion problem is a free fall problem. • Horizontal and vertical parts of the motion are independent • To solve a projectile motion problem, break up the motion into x and y components. Kinematics Equations for uniformly accelerated motion Horizontal (ax=0) x v0 xt Vertical (FREE FALL) y v yt v y v0 y a yt vy 1 2 (v0 y v y ) y v0 yt v v 2 y 2 0y 1 2 a yt 2 2a yy Remember… To work projectile problems… • …resolve the initial, launch velocity into horizontal and vertical components. Vo V0x = Vo cos V0y = Vo sin Some Common Projectile Motion Problems Horizontal Launch v0 When object is launched horizontally: v0y = 0 x, Range Angle Launch on Level Ground When object returns to the same height from which it was launched: v0 x, Range y = 0 Projectile Launched from Any Angle y vt vty v0 v0y vx vx vx vx ymax = height a = constant along y At max height, vy = 0 vx vx vx v = constant along x xmax = range vfy = -v0y at landing Initial Horizontal velocity v 0 x v x v 0 cos Initial Vertical velocity (determines air time) v 0 y v 0 sin x vf PM simulation EXAMPLE A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 67.1 m from the base of the bluff. She launches a horizontal shot that lands in the hole on the fly. The gallery erupts in cheers. a) How long was the ball in the air? b) What was the ball’s initial velocity? c) What was the ball’s impact velocity (velocity right before landing) y x v x v0 v0 y 0 x 67 . 1 m vy t a y 9 .8 m / s y 12 . 5 m 12.5 m t = 1.6 s t v0 = 41.9 m/s vf = 44.7 m/s, 20.50 below the horizontal 67.1 m 2 EXAMPLE In the circus, a clown is launched from a cannon at 40 m/s, 60o from the horizontal. Where should the other clowns hold the net so that the projectile clown lands unharmed (at the same level)? t = 7.07 s x = 141 m v 0 40 m / s 60 0 EXAMPLE A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top if it is to land on level ground below, 90.0 m from the base of the cliff where the cameras are? What is its final velocity, right before landing? vo = 28.5 m/s v0 = ? 50.0 m x y v0 y 0 vx ? x 90 m v y 2 t a y 9 .8 m / s y B 50 . 0 m t 90.0 m EXAMPLE What is its final velocity, right before landing? vf = 42.3 m/s, -47.70 from x-axis x y v0 y 0 v x 28 . 5 m / s vy x 90 m 2 t a y 9 .8 m / s y B 50 . 0 m t 50.0 m vf = ? 90.0 m EXAMPLE What horizontal firing (muzzle) velocity splashes the cannonball into the pond? x - component v x v0 x 95 m t v0 y - component v0 y 0 vy ay 40 m t = 2.86 s v0 = 33.2 m/s 95 m y 40 m t EXAMPLE Where does the ball land? How far has the truck traveled in that time? Ball x v x 30 m / s x t v = 10 m/s y v 0 y 10 m / s vy a y 9 .8 m / s 2 y 0 v = 30 m/s t t = 2.0 s x = 60 m B1 B2 http://zamestnanci.fai.utb.cz/~schauer/PhET1.0/simulations/simsa79a.html?sim=Projectile _Motion Projectile Launched from Any Angle Horizontal velocity v 0 x v 0 cos Vertical velocity v 0 y v 0 sin When starting and landing heights are the same: Time of flight Range t 2v0 y 2 v 0 sin ay ay 2 v 0 cos sin 2 x ay v 0 sin 2 Max height y max 2a y v 0 sin 2 2 ay 2 Max at 450 All depend on angle and initial velocity Remember… To work projectile problems… • …resolve the initial velocity into components. Vo Vo,x = Vo cos Vo,y = Vo sin DO NOW Sketch the following motion graphs for a horizontally launched projectile. x y t t vx vy t t ax ay t t DO NOW A fire hose held near the ground shoots water at a speed of 6.6 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there 2 different angles? = 13.70, 76.30 v 0 6 .6 m / s x 2 .0 m EXAMPLE A professional soccer player is 25.8 m from the goal and kicks a hard shot at ground level. The ball hits the cross bar on its way down, 2.44 m above the ground, 1.98 s after being kicked. What was the initial velocity of the ball (find the x- and y- components)? x v x v 0 cos x 25 . 8 m t 1 . 98 s y vx = 13.0 m/s v0y = 11.13 m/s v0 = 17.1 m/s, 40o v 0 y v 0 sin vy a y 9 .8 m / s 2 y 2 . 44 m v0 ? t 1 . 98 s 2.44 m 25.8 m EXAMPLE A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top if it is to land on level ground below, 90.0 m from the base of the cliff where the cameras are? What is its final velocity, right before landing? vo = 28.5 m/s v0 = ? 50.0 m x y v0 y 0 vx ? x 90 m v y 2 t a y 9 .8 m / s y B 50 . 0 m t 90.0 m EXAMPLE What is its final velocity, right before landing? vf = 42.3 m/s, -47.70 from x-axis x y v0 y 0 v x 28 . 5 m / s vy x 90 m 2 t a y 9 .8 m / s y B 50 . 0 m t 50.0 m vf = ? 90.0 m EXAMPLE A player kicks a football from ground level with an initial velocity of 27.0 m/s 300 above the horizontal. Assume air resistance is negligible. a) Find the ball’s hang time b) Find the ball’s maximum height c) Find the ball’s range d) Find the impact velocity Player then kicks the ball with the same speed but at a 60 degree angle. Find all the above v 0 27 . 0 30 0 300 a) t = 2.76 s b) ymax = 9.30m c) range=64.5m d) vf=27m/s -300 600 a) t = 4.78 s b) ymax = 27.9m c) range=64.5m d) vf=27m/s -600 EXAMPLE Agent Tim flying a constant 185 km/hr horizontally in a low flying helicopter, wants to drop a small explosive onto a master criminal’s automobile, traveling 145 km/hr on a level highway 88.0 m below. At what angle (with the horizontal) should the car be in his sights when the bomb is released? 62.1o 185 km/h 88.0 m 145 km/h DO NOW Frustrated with the book you are reading, you open the second story classroom window and violently hurl your book out the window with a velocity of 18 m/s at an angle of 35o above the horizontal. If the launch point is 6 m above the ground, a) how far from the building will the book hit the parking lot? b) What is the final velocity of the book right before it hits the ground? x = 37.9 m vf = 21 m/s, 45.40 below the x-axis “Projectile motion and parabolas”: Football kick, projectile motion http://www.nbclearn.com/portal/site/learn/nfl /cuecard/50974/ Wile E. Coyote Physics Problem The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of 15 m/s^2. The coyote starts off at rest 70m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. A) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have in order to reach the cliff before the coyote. ____m/s B) If the cliff is 100m above the base of a canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at 15m/s^2. ____m from the base I don't need answers, but I do need help on how to get there.